Let $f: A \to B$. Let $f*$ be the inverse relation, i.e. \begin{equation*} f* = \{(y,x) \in B \times A \mid f(x)=y \}. \end{equation*} Show that if $f* : B \to A$ is a function, then $f^{-1} = f*$.


Let $f*: B \to A$ be a function. Then, $f$ is bijective and hence $f$ is invertible, i.e. $f$ have an inverse, say $f^{-1}$. It is clear that $f^{-1} \circ f = i_A$ and $f \circ f* = i_B$ where $i$ is the identity function. Hence, \begin{equation*} f^{-1} = f^{-1} \circ i_B = f^{-1} \circ (f \circ f*) = (f^{-1} \circ f) \circ f* = i_A \circ f* = f*. \end{equation*} Thus, $f^{-1} = f$.

Is the above correct?

  • 1
    $\begingroup$ Since $i = f^{-1}\circ f : A \to A$ is the identity function on $A$, how is defined $f^{-1}\circ i$? Also, $f \circ f^*$ is the identity function on $B$, it is not the same as $i$. $\endgroup$
    – azif00
    Commented Jan 6, 2021 at 7:28
  • $\begingroup$ I assume your defintion of $f^{-1}$ is "any left-inverse"? And instead of "It is clear that", you may argue more explicitly where you use the fact that $f^*$ is a function. $\endgroup$ Commented Jan 6, 2021 at 7:29
  • $\begingroup$ Suggestion for improvement: Compute the relation(!) compositions $f^*\circ f\subseteq A\times A$ and $f\circ f^*\subseteq B\times B$ and see what happens when both $f$ and $f^*$ are functions $\endgroup$ Commented Jan 6, 2021 at 7:32
  • $\begingroup$ The question is missing a proper definition of $f^{-1}$. I will not settle for "the usual definition of the inverse". $\endgroup$
    – user65203
    Commented Jan 6, 2021 at 7:52
  • $\begingroup$ What is the connection of the title with the question ? $\endgroup$
    – user65203
    Commented Jan 6, 2021 at 7:54

1 Answer 1


First, suppose that $f*$ is a function. That means that each for each $y \in B, \exists!x \in A$ such that $f^*(y)=x.$ Then $(f^* \circ f)(x) = f^*(f(x)) = f^*(y)=x,$ and similarly, $(f \circ f^*)(y) =f(f^*(y))= f(x)=y$, so $f*$ is the inverse of $f$. Or perhaps I should say $f*$ is an inverse of $f$, since you've also been tasked with proving the uniqueness of the inverse.

To prove that the inverse of f is unique, suppose that f has two inverses, g and h. We need to show that g=h. Since g is the inverse of f, that means $g∘f=id_A$ and $f∘h=id_B$. Then $g=g∘id_B=g∘(f∘h)=(g∘f)∘h=id_A∘h=h$.

  • $\begingroup$ but, how to prove $f^{-1} = f*$. $\endgroup$
    – lap lapan
    Commented Jan 6, 2021 at 10:35
  • $\begingroup$ Is my answer above correct? $\endgroup$
    – lap lapan
    Commented Jan 6, 2021 at 10:41
  • $\begingroup$ Yes, the basic proof that $f*$ is an inverse of $f$ is correct, you just needed the uniqueness part. And at the end, did you mean to put "Thus, $f^{-1} = f^*$"? $\endgroup$
    – Hank Igoe
    Commented Jan 6, 2021 at 10:47
  • $\begingroup$ N1 bro, thanks for give another views. $\endgroup$
    – lap lapan
    Commented Jan 6, 2021 at 10:49
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    $\begingroup$ You could use either one since they are both correct. So you could just pick whichever one is more similar in style to the way the professor does examples in class. $\endgroup$
    – Hank Igoe
    Commented Jan 6, 2021 at 11:07

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